What is the de Broglie wavelength in cm of a 12.4g hummingbird flying 1.2x102mph?

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{mv}$$

where:

- λ is the de Broglie wavelength in meters (m)

- h is the Planck constant ( 6.626 x 10-34 Js)

- m is the mass of the object in kilograms (kg)

- v is the velocity of the object in meters per second (m/s)

First, convert the mass to kilograms:

$$m = 12.4g (\frac{1 kg}{1000g})= 0.0124 kg$$

Then, convert the velocity to meters per second:

$$v = (1.2 \times 10^2 mph) (\frac{1609.344 m}{1 mi})(\frac{1 h}{3600 s}) = 53.6448 m/s$$

Now, we can plug these values into the de Broglie wavelength equation:

$$\lambda = \frac{6.626 \times 10^{-34} Js}{(0.0124 kg)(53.6448 m/s)} = 1.04 \times 10^{-34} m$$

Finally, convert the wavelength to centimeters:

$$1.04 \times 10^{-34} m(\frac{100 cm}{1 m}) = \boxed{1.04 \times 10^{-32}\ cm}$$